The volume occupied by 11 grams of $C O_{2}$ gas at STP is:

5.6 l

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11.2 l

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22.4 l

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44.8 l

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Solution

The correct option is A $5.6 l$
As we know,
One mole of a gaseous substance occupies $22.4 l i t r e$ .
mole of $C O_{2} = 11 / 44 = 1 / 4$ mole
volume occupied $= \frac{22.4}{4} = 5.6 l i t r e$

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